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Units and Measurements question

2011 · Shift 1 · Q47
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Units and Measurements question

2011 · Shift 1 · Q47

JEE AdvancedPhysicsUnits and MeasurementsMCQ+3 / −0.75
A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let 'N' be the number density of free electrons, each of mass 'm'. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency 'ωp{\omega _p}ωp​' which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency ω\omegaω, where a part of the energy is absorbed and a part of it is reflected. As ω\omegaω approaches ωp{\omega _p}ωp​ all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals. Taking the electronic charge as 'e' and the permittivity as ′ε0′'{\varepsilon _0}'′ε0​′. Use dimensional analysis to determine the correct expression for ωp{\omega _p}ωp​.
  1. A
    Nemε0\sqrt {{{Ne} \over {m{\varepsilon _0}}}}mε0​Ne​​
  2. B
    mε0Ne\sqrt {{{m{\varepsilon _0}} \over {Ne}}}Nemε0​​​
  3. C
    Ne2mε0\sqrt {{{N{e^2}} \over {m{\varepsilon _0}}}}mε0​Ne2​​
  4. D
    mε0Ne2\sqrt {{{m{\varepsilon _0}} \over {N{e^2}}}}Ne2mε0​​​
View written solutionFree

Correct answer: C

  1. We need an expression for plasma angular frequency ωp\omega_pωp​.

    Since it is an angular frequency, its dimension is [ωp]=T−1.[\omega_p] = T^{-1}.[ωp​]=T−1.

  2. Assume ωp∝Naebmcε0d.\omega_p \propto N^a e^b m^c \varepsilon_0^d.ωp​∝Naebmcε0d​.

    So, [ωp]=[N]a[e]b[m]c[ε0]d.[\omega_p] = [N]^a [e]^b [m]^c [\varepsilon_0]^d.[ωp​]=[N]a[e]b[m]c[ε0​]d.

  3. Write dimensions of each quantity:

    • Number density: [N]=L−3[N] = L^{-3}[N]=L−3

    • Charge: [e]=IT[e] = IT[e]=IT

    • Mass: [m]=M[m] = M[m]=M

    • Permittivity:

      From Coulomb's law, F=14πε0q2r2F = \frac{1}{4\pi \varepsilon_0}\frac{q^2}{r^2}F=4πε0​1​r2q2​ hence [ε0]=q2Fr2.[\varepsilon_0] = \frac{q^2}{F r^2}.[ε0​]=Fr2q2​.

      Now,

      \quad [F] = MLT^{-2}, \quad [r^2] = L^2,$$ so $$[\varepsilon_0] = \frac{I^2T^2}{(MLT^{-2})(L^2)} = M^{-1}L^{-3}T^4I^2.$$

  4. Therefore, [ωp]=(L−3)a(IT)b(M)c(M−1L−3T4I2)d.[\omega_p] = (L^{-3})^a (IT)^b (M)^c (M^{-1}L^{-3}T^4I^2)^d.[ωp​]=(L−3)a(IT)b(M)c(M−1L−3T4I2)d.

    Collecting powers: [ωp]=Mc−dL−3a−3dTb+4dIb+2d.[\omega_p] = M^{c-d} L^{-3a-3d} T^{b+4d} I^{b+2d}.[ωp​]=Mc−dL−3a−3dTb+4dIb+2d.

  5. Compare with T−1T^{-1}T−1: M0L0T−1I0.M^0L^0T^{-1}I^0.M0L0T−1I0.

    So, \begin{align*} c-d &= 0, \ -3a-3d &= 0, \ b+4d &= -1, \ b+2d &= 0. \end{align*}

  6. Solve these equations:

    From b+2d=0b+2d=0b+2d=0, b=−2d.b=-2d.b=−2d.

    Put into b+4d=−1b+4d=-1b+4d=−1: −2d+4d=−1⇒2d=−1⇒d=−12.-2d+4d=-1 \Rightarrow 2d=-1 \Rightarrow d=-\frac12.−2d+4d=−1⇒2d=−1⇒d=−21​.

    Hence, b=1.b=1.b=1.

    From c−d=0c-d=0c−d=0, c=d=−12.c=d=-\frac12.c=d=−21​.

    From −3a−3d=0-3a-3d=0−3a−3d=0, a=−d=12.a=-d=\frac12.a=−d=21​.

  7. Therefore,

    = \sqrt{\frac{Ne^2}{m\varepsilon_0}}.$$

  8. Match with options: ωp=Ne2mε0\boxed{\omega_p = \sqrt{\frac{Ne^2}{m\varepsilon_0}}}ωp​=mε0​Ne2​​​ which is option C\boxed{\text{C}}C​.

  9. Verification with stored answer:

    Stored correct answer = C.

    Our derived answer also = C, so they agree.

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